Question

The sum of the squares of three distinct real numbers which are in g.p is s^2 . if their sum is as ,show that a^2 belongs (1/3,1)U(1,3)

Answer 1

let b,br,br 2 are 3 number
so s 2 = (b) 2 +(br) 2 +(br 2 )2 ..............1
and as= b +br +br 2 ..................2
square 2 eqation and divide it by
equation 1
a 2 =(b +br +br 2 )2 /((b) 2 +(br) 2 +(br 2 )
2 )
=(1 +r +r 2 ) 2 /( 1+r 2 +r 4 )
=(1 +r +r 2 )/(1 -r +r 2 )
rearrange
r 2 (a 2-1) -r(a 2 +1) +(a 2 -1)=0
discrimnent must be greter than zero
(a 2 +1) 2-4(a 2-1) 2 >=0
or
(3a 2-1)(3-a 2)>=0
or
(a-√3)(a+√3)(a-1/3)(a+1/3)<=0
find a from this enequality
reject those value of a for which r=0 or
r=1(since numbers are in gp)
HOPE IT HELPS